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Ap Calculus B.C. This involves improper integrals. I need help so give me a detailed explanation with steps.

Ap Calculus B.C. This involves improper integrals. I need help so give me a detailed-example-1
User Smorka
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\int ^(-1)_(-\infty)e^(5t)dt

The first thing we are going to do is a substitution:


\begin{gathered} u=5t \\ du=5dt\to dt=(1)/(5)du \end{gathered}

We replace the substitution in the integral and solve the integral


\begin{gathered} \int ^(-1)_(-\infty)e^u\cdot((1)/(5)\cdot du) \\ (1)/(5)\int ^(-1)_(-\infty)e^u\cdot\mathrm{d}u \end{gathered}

We know that the integral of an exponential is the same exponential, so:


(1)/(5)\int ^(-1)_(\infty)e^u\cdot\mathrm{d}u=(1)/(5)e^u

To evaluate the answer we must recover the substitution made from the beginning:


(1)/(5)e^u=(1)/(5)e^(5t)

Now, we evaluate in the interval from negative infinity to negative 1


\begin{gathered} \int ^(-1)_(-\infty)e^(5t)dt=(1)/(5)e^(5t) \\ \int ^(-1)_(-\infty)e^(5t)dt=(1)/(5)e^(5(-1))-(1)/(5)e^(5(-\infty)) \\ \int ^(-1)_(-\infty)e^(5t)dt=(1)/(5)e^(-5)-0 \end{gathered}
\int ^(-1)_(-\infty)e^(5t)dt=(1)/(5)e^(-5)=0.00135

This integral is convergent

User Sajanyamaha
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